# Does a finite mass particle pass a black hole event horizon in a finite time for an outside observer?

When we talk about a particle taking an infinite amount of time for it to cross the event horizon in the external observer's point of view, we assume that the particle follows a geodesic and does not affect the metric. However, if we assume the that the mass of the particle is not zero, it perturbs the metric a little. So, in theory, would the particle be able to enter the horizon in a finite time (for the external observer)?

I suspect the answer is yes because we have already simulations of Black hole mergers and they collide in a finite time span. We can think of one of the Black holes as a very massive particle since both particles and Black holes are point masses. Another example is that of a collapsing dust. We see again mass entering the horizon in a finite time span if the infalling mass can alter the geometry of space.

In addition, if one needed to calculate the trajectory of the infalling particle, how would he go about doing it? If we consider a point particle, there would be a singularity at the position of the particle and, as a result, we don't have any valid geodesics at the location of the particle.

• Black holes take an infinite time to merge. However they approach the final state so quickly that within a very short time they are indistinguishable from the fully merged state. Commented Apr 30, 2021 at 16:53
• I see, the problem is defining when the particle can be considered to be completely inside the black hole. But what if I defined the particle being completely in the black hole as the singularity of the particle entering the event horizon of the black hole? (Even though it does not approach the kerr metric) Commented May 2, 2021 at 17:14

There is a simple argument to show that a particle must take a infinite (coordinate) time to reach to reach a horizon, and this applies to all particles regardless of their mass. The argument is simply that in GR geodesic motion is time symmetric i.e. the geodesic can be traversed in either direction.

If a particle can reach a horizon in a finite time we can reverse the motion and the particle will escape the horizon in a finite time. But by definition a horizon is a surface from which there is no escape so no point reached by a particle in a finite time can be part of an event horizon.

A few quick notes:

• in the above time means coordinate time as measured by an observer outside the horizon

• this ignores black hole evaporation due to Hawking radiation

I don't think it matters.

The only sense in which test particles take forever to cross the event horizon "from the perspective of" someone outside is that news of the crossing, conveyed by light or any other causal signal, takes an unbounded time to reach them. News of events just before the crossing reaches them quickly, though.

If you replace the test particle by a small black hole, you'll still never get news of the merging of the event horizon(s), but news of the shape of spacetime just outside the horizon(s) reaches you quickly.

In videos of merging black hole horizons, like this one, they are showing you slices through the manifold they calculated. There's a large element of arbitrary choice in the way they slice it up. They may choose a slicing that in some way reflects the gravitational-wave signal that would be seen from a large distance by an experiment like LIGO, but what LIGO really sees is not the horizon but events just outside it.

A "point" particle in general relativity needn't be a naked singularity; it could be a small black hole, which avoids the infinities that are associated with point particles even in flat spacetime. Unfortunately light electrically charged elementary particles like electrons should be naked singularities according to a naive calculation, and dealing with that is an open problem. See black hole electron on Wikipedia.